First Sylow Theorem: Group of Order ##p^k## & Cyclic Groups

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SUMMARY

The first Sylow theorem states that for a group G of order ##p^m n##, where gcd(n, p) = 1, there exists a subgroup H of G with order ##p^m##. In the case of a group G with order ##p^k##, the only subgroup of that order is G itself, indicating that G is indeed cyclic when p is prime. This conclusion directly links the structure of groups of prime power order to their cyclic nature, confirming that all such groups are cyclic.

PREREQUISITES
  • Understanding of group theory fundamentals
  • Familiarity with Sylow theorems
  • Knowledge of cyclic groups and their properties
  • Basic concepts of finite Abelian groups
NEXT STEPS
  • Study the implications of the second and third Sylow theorems
  • Explore examples of finite Abelian groups that are not cyclic
  • Learn about the classification of finite groups
  • Investigate the structure of groups of order ##p^k## for various primes p
USEFUL FOR

Mathematics students, group theorists, and anyone studying the properties of finite groups and their classifications.

Silviu
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Hello! I am a bit confused about the first Sylow theorem. So it says that if you have a group of order ##p^mn##, with gcd(n,p)=1, you must have a subgroup H of G of order ##p^m##. So, if I have a group G of order ##p^k##, there is only one subgroup of G of order ##p^k## which is G itself. Does this means that any group of order ##p^k##, with p prime is cyclic?

Thank you!
 
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Which other finite Abelian groups do you know which are not cyclic?
 

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